How to find the limit for lim x -> 0 ( 1/(xsqrtx) * int_0^sqrtx cos(pi/2*e^(t^2))dt) with l'Hospital?

1 Answer
Jun 11, 2018

- pi/6

Explanation:

lim_(x -> 0) ( int_0^sqrtx cos(pi/2*e^(t^2))dt)/ (xsqrtx)

That is 0/0 indeterminate so we use LHR and differentiate top and bottom wrt x.

For the top use the Liebnitz Integral Rule, which applies here as:

  • bb( d/dx ( int_0^(v(x)) f(s) \ ds ) = f(v(x)) * v^'(x) )

= lim_(x -> 0) ( cos(pi/2*e^(x))* 1/(2 sqrtx))/ (3/2 sqrtx)

= lim_(x -> 0) ( cos(pi/2*e^(x)) )/ (3 x)

That is still 0/0 indeterminate so we go again with LHR.

= lim_(x -> 0) ( pi/2 e^x * - sin(pi/2*e^(x)) )/ (3 ) = ((pi/2)(-sin (pi/2)) ) /3= - pi /6